混淆静态int并在printf中调用它们

Question

Please test this code and give me your answers :

#include <stdio.h>

int func() {
static int n = 0;
n++;
return n;
}

int main() {
    /*int first = func();
    int second = func();*/
    printf(" first call : %d \n second call : %d ",func(),func());
    return 0;
}

Logically it should print 1 and 2 but it is printing 2 and 1 . If you Un-Comment the comments and print the variables "first" and "second" , the problem is solved! What is happening?

thank you already!

Answer 1

The order in which a function call's arguments are evaluated is unspecified, ie, the compiler is free to make the two func() calls in any order before passing the return values to printf . If you first assign the results to variables, obviously you get to decide in which order they are used.

Answer 2

The order in which the parameters to a function are passed is not defined in the standard, and is determined by the calling convention used by the compiler. I think in your case, cdecl calling convention (which many C compilers use for x86 architecture) is used in which arguments in a function get evaluated from right to left.

Answer 3

because while calling the function it takes right associative.. ie, the last one in printf gets called first then the before one.. thats why it is printing 2 then 1. try using two print statements like: printf("First call: %d\\n",func()); printf("second call: %d\\n",func());